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ivolga24 [154]
2 years ago
15

Why can't all mixtures be classified as solutions?

Chemistry
1 answer:
shusha [124]2 years ago
7 0
D. Not all mixtures are heterogeneous
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sasho [114]
Answer is attached  below ( sorry for mistake )

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3 years ago
All of the food that we eat, liquids that we drink and medications that we take are
kirill115 [55]

A) Chemicals

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3 0
3 years ago
Mg + 2HCl → MgCl2 + H2
deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

+

2HCl(aq)

→

MgCl

2

(

aq

)

+ H

2

(

g

)

Determine Moles of Magnesium

Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

g Mg

×

1

mol Mg

24.3050

g Mg

=

0.200 mol Mg

Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

3

to

100 mL

and then to

0.1 L

.

1 dm

3

=

1 L

Convert

2.00 mol/dm

3

to

2.00 mol/L

Multiply

0.1

L

times

2.00 mol/L

.

100

cm

3

×

1

mL

1

cm

3

×

1

L

1000

mL

=

0.1 L HCl

2.00 mol/dm

3

=

2.00 mol/L

0.1

L

×

2.00

mol

1

L

=

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

2.01588 g/mol

0.200

mol Mg

×

1

mol H

2

1

mol Mg

×

2.01588

g H

2

1

mol H

2

=

0.403 g H

2

0.200

mol HCl

×

1

mol H

2

2

mol HCl

×

2.01588

g H

2

1

mol H

2

=

0.202 g H

2

The limiting reactant is

HCl

, which will produce

0.202 g H

2

under the stated conditions.

pls mark as brainliest ans

7 0
2 years ago
Read 2 more answers
What is the effect of temperature on the solubility of a solid in a liquid
krok68 [10]
For many solids<span> dissolved in </span>liquid<span> water, the </span>solubility <span>increases with </span>temperature<span>.</span>
7 0
3 years ago
Read 2 more answers
Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What
coldgirl [10]

Answer:

\rm_{90}^{231}\text{Th}

Explanation:

The unbalanced nuclear equation is

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + X

Let's write X as a nuclear symbol.

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + _{Z}^{A}\text{X}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

 92 = 2 + Z , so  Z =   92 - 2 =  90

And your nuclear equation becomes

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} +\, _{90}^{231}\text{X}

Element 90 is thorium, so  

\rm X = _{90}^{231}\text{Th}

7 0
3 years ago
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